Let $(P_n(z))_{n\geq 1}$ be the sequence of polynomials defined by
$$ \begin{array}{lcl} P_0(z) &=& 0 \\ P_{n+1}(z) &=& P_n(z)^2-z \end{array} $$
It is related to the Mandelbrot set. I have checked for $n\leq 6$ that
Conjecture. When $z\in{\mathbb C}, |z|\geq 2$ and $n\geq 0$, we have $|P_n(z)|\geq P_n(|z|)$.
Can anyone prove or disprove this conjecture ?
What I did: put $r=|z|$. The number $u=\frac{z}{r}$ has modulus one. Unless $u=-1$, there is a $t\in{\mathbb R}$ such that $u=\frac{(1-t^2)+2t i}{1+t^2}$. Expanding $|P_n(z)|-P_n(|z|)=\Bigg|P_n\Big(r\big(\frac{(1-t^2)+2t i}{1+t^2}\big)\Big)\Bigg|-P_n(r)$, we find that it is of the form $\frac{A_n(r,t)}{(1+t^2)^{2^{n-1}}}$ where $A_n(r,t)$ is a polynomial. Writing $A_n(r,t)=\sum_{k=0}^N B_k(r) t^k$ where $N$ is the degree of $A_n$ with respect to $t$, it turns out that for $n\leq 6$, each $B_k(r)$ is nonnegative when $r\geq 2$.
Define a function of $x$ by $$ \phi_z(x) := x^2 - z. $$
First, we prove that $|\phi_z(x)| \ge \phi_{|z|}(|x|)$. Indeed, $$ |\phi_z(x)| = |x^2 - z| \ge ||x|^2 - |z|| \ge \phi_{|z|}(|x|) $$ by the second triangle inequality. Now we proceed to the desired inequality. Note the formula $$ P_{n+1}(z) = \phi_z(P_n(z)). $$ Hence, suppose we have proven the statement for $n$. Then $$ |P_{n+1}(z)| = |\phi_z(P_n(z))| \ge \phi_{|z|}(|P_n(z)|) \ge \phi_{|z|}(P_n(|z|)) = P_{n+1}(|z|), $$ where the last inequality follows from the induction hypothesis and the fact that $\phi_{|z|}$ increases monotonically on the positive reals - this follows immediately from $$ \frac{d}{dx} \phi_z(x) = 2x. $$ Indeed, we are restricted to the positive real numbers because $P_n(|z|) \ge |z| \ge 0$ for $|z| \ge 2$. Indeed, $$ P_n(|z|) = P_{n-1}(|z|)^2 - |z| \overset{\text{induction}}{\ge} |z|(|z|-1) \ge |z|. $$